If $R$ is a Dedekind domain and $I, J \triangleleft R$, then $R^d \oplus I \simeq R^{d'} \oplus J$ implies $d = d'$ and $I \simeq J$ I'm trying to figure out the following observation I have written on my notes on Dedekind domains,

If $R$ is a Dedekind domain, $I, J$ two integral ideals and $d, d' \in \mathbb{N}$, such that $R^d \oplus I \simeq R^{d'} \oplus J$, then we have that $d = d'$ and $I \simeq J$.

It seems quite simple, but given my lack of familiarity with abstract algebra, I haven't made much progress. I've tried composing the given isomorphism $\phi: R^d \oplus I \longrightarrow R^{d'} \oplus J$ with the projections to both coordinates to get each summand isomorphic to $R^d \oplus I$ divided by the preimages of $R^{d'} \oplus 0$ and $0 \oplus J$ respectively, but it is not obvious to me that (if) we indeed have $\phi(R^d \oplus 0) = R^{d'} \oplus 0$ and $ \phi(0 \oplus I) = 0 \oplus J$. 
Would you mind providing any hints? Thanks in advance.
 A: To understand the answer you need to know some properties of Dedekind domains. 
First of all, any finitely generated torsion-free module over a Dedekind domain has a well defined rank, which can be either defined as the rank of the base change to the quotient field, or the maximum number of linearly independent elements (but not the minimum number of generators). Ideals, and more generally, fractional ideals, have rank 1 (but need more than 1 generator if they are not principal): in fact, fractional ideals are the (torsion-free) rank 1 $R$-modules. 
Two isomorphic finitely generated torsion-free modules (=projective) have the same rank, but the reverse is not true. For example, ideals are isomorphic (as $R$-modules) if and only if they represent the same class in the class group. 
Now, for any finitely generated torsion-free $R$-module $M$ of rank $d$, its wedge product $\lambda(M):=\Lambda^dM$ (this notation is not standard, I am not sure if there is a standard notation) is a torsion-free $R$-module of rank 1, so a fractional ideal. If $M=R^{d-1}\oplus I$, then $\lambda(M)\cong I$, and hence $M=R^{d-1}\oplus I\cong R^{d-1}\oplus I'=M'$, then $\lambda(M)\cong \lambda(M')$, so $I\cong I'$. 
In fact, something more is true: if $M$ and $M'$ are torsion-free finitely generated $R$-modules, they are isomorphic if and only if $rank_R(M)=rank_R(M')$ and $\lambda(M)\cong \lambda(M')$, or, equivalently, they represent the same element in the class group of $K$. 
